Question

If \(x = (9 + 4\sqrt{5})^{48} = [x] + f\), where \([x]\) is the integral part of \(x\) and \(f\) is a fraction, then \(x(1-f)\) equals:

• A. 1
• B. Less than 1
• C. More than 1
• D. Between 1 and 2
• E. None of the above

Hint:

For expressions involving \((a+b\sqrt{c})^n\), always think of its conjugate \((a-b\sqrt{c})^n\). Their sum often becomes an integer and their product simplifies to a power of an integer. This trick avoids complex binomial expansion.

Answer 1

The Correct Option is A

Step 1: Define the given terms.
We are told: \[ x = (9 + 4\sqrt{5})^{48} = [x] + f \] where \([x]\) is the integer part and \(f\) is the fractional part with \(0<f<1\).

Step 2: Introduce the conjugate.
Let \[ y = (9 - 4\sqrt{5})^{48} \] Notice that \(9 - 4\sqrt{5} \approx 0.055<1\). Raising it to the power 48 makes it extremely small, so \(0<y<1\).

Step 3: Use integer property.
From binomial expansion: \[ (9 + 4\sqrt{5})^{48} + (9 - 4\sqrt{5})^{48} \] is an integer, because irrational terms cancel out. Thus, \[ x + y = \text{integer.} \]

Step 4: Relating fractional part.
Since \(x = [x] + f\) and \(x + y\) is integer, we can write: \[ [x] = x + y - 1 \] So, \[ f = 1 - y \]

Step 5: Expression for \(x(1-f)\).
We need: \[ x(1-f) = x \cdot (1-f) = x \cdot y \]

Step 6: Simplify product.
\[ x \cdot y = (9 + 4\sqrt{5})^{48}(9 - 4\sqrt{5})^{48} \] \[ = \big[(9)^2 - (4\sqrt{5})^2\big]^{48} \] \[ = (81 - 80)^{48} = 1^{48} = 1 \]

Step 7: Conclude.
Thus, the exact value of \(x(1-f)\) is: \[ \boxed{1} \]